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KonstantinChe [14]

9 months ago

4

The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t

. [5 marks] ii.Does the object ever stop changing? [5 marks] iii.When is the object moving to the right and when is the object moving to the left?

1 answer:

kozerog [31]9 months ago

0 0

Answer:

1.) V = 12t^3 - 120t^2 + 252t

2.) No

3.) At positive V and negative V

Explanation: Given that the position of an object at any time t is s(t)=3t4−40t3+126t2−9.

1.) To determine the velocity V, we will differenciate the equation above with respect to t.

V = ds/dt

ds/dt = 12t^3 - 120t^2 + 252t

Therefore,

V = 12t^3 - 120t^2 + 252t

2.) The equation above depict an exponential equation which proves that the object never stops changing.

3.) The object moves to the right at positive velocity V and moves to the left at negative velocity V

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An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws tw

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Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

$y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{1}$ - Final height of the first stone, measured in meters.

$y_{1,o}$ - Initial height of the first stone, measured in meters.

$v_{1,o}$ - Initial speed of the first stone, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{1,o} = 47\,m$ , $y_{1} = 0\,m$ , $v_{1,o} = -2.12\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the following second-order polynomial is built:

$-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0$

Roots of the polynomial are, respectively:

$t_{1} \approx 2.866\,s$ and $t_{2}\approx -3.291\,s$

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

$y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}$

$y_{2}$ - Final height of the second stone, measured in meters.

$y_{2,o}$ - Initial height of the second stone, measured in meters.

$v_{2,o}$ - Initial speed of the second stone, measured in meters per second.

$t$ - Time, measured in seconds.

$t_{o}$ - Initial absolute time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{2,o} = 47\,m$ , $y_{2} = 0\,m$ , $t_{o} = 1\,s$ , $t = 2.866\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the following expression is constructed and the initial speed of the second stone is:

$1.866\cdot v_{2,o}+29.926 = 0$

$v_{2,o} = -16.038\,\frac{m}{s}$

The initial velocity of the second stone is -16.038 meters per second.

c) The final speed of each stone is determined by the following expressions:

First stone

$v_{1} = v_{1,o} + g \cdot t$

Second stone

$v_{2} = v_{2,o} + g\cdot (t-t_{o})$

Where:

$v_{1,o}, v_{1}$ - Initial and final velocities of the first stone, measured in meters per second.

$v_{2,o}, v_{2}$ - Initial and final velocities of the second stone, measured in meters per second.

If $v_{1,o} = -2.12\,\frac{m}{s}$ and $v_{2,o} = -16.038\,\frac{m}{s}$ , the final speeds of both stones are:

First stone

$v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)$

$v_{1} = -30.227\,\frac{m}{s}$

Second stone

$v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)$

$v_{2} = -34.338\,\frac{m}{s}$

The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

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